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Subnetting question

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rprudent
Level 1
Level 1

‎07-08-2010 06:31 PM - edited ‎03-06-2019 11:57 AM

Hi all,

Can someone explain why these 2 ip addresses are correct and show how the ip's where achieved. Thanks

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michael.leblanc
Level 4
Level 4

‎07-13-2010 08:58 AM

rprudent:

The easiest way to get a handle on these, is with a thought process similar to the following:

  • With a mask of 255.255.254.0, it is the third octet that distinguishes adjacent Networks IDs.
  • If you subtract the value of the third octet (254) from 256 (0 - 255  represents 256 possible values), you will get 2.
  • This value (2) represents the increment between Network IDs.

E.g.: 113.10.4.0 255.255.254.0 would yield the following Network IDs:

113.10.0.0
113.10.2.0
113.10.4.0
113.10.6.0
.

. etc.
113.10.252.0
113.10.254.0


Note: that the third octet is incrementing by 2.

  • The address that follows a Network ID (e.g.: 113.10.4.0), is the first assignable host address (e.g.: 113.10.4.1).
  • The address that precedes the "next" Network ID (e.g.: 113.10. 6.0), would be the broadcast address (e.g.: 113.10.5.255).
  • The address that precedes the broadcast address (e.g.: 113.10.5.255), would be the last assignable host address (e.g.: 113.10.5.254).

Answer Assessment


A.     113.10.4.0 255.255.254.0 defines an address space of 113.10.4.0 - 113.10.5.255

The network ID would be: 113.10.4.0
The broadcast address would be 113.10.5.255
The host address range would be 113.10.4.1 - 113.10.5.254, which is not inclusive of 113.10.4.0

Note: The address in question, happens to be a Network ID.


B.     186.54.3.0 255.255.254.0 defines an address space of 186.54.2.0 - 186.54.3.255

The network ID would be: 186.54.2.0
The broadcast address would be 186.54.3.255
The host address range would be 186.54.2.1 - 186.54.3.254, which is inclusive of 186.54.3.0

Note: The address in question, is host assignable.


C.     175.33.3.255 255.255.254.0 defines an address space of 175.33.2.0 - 175.33.3.255

The network ID would be: 175.33.2.0
The broadcast address would be 175.33.3.255
The host address range would be 175.33.2.1 - 175.33.3.254, which is not inclusive of 175.33.3.255

Note: The address in question, happens to be the broadcast address.


D.     26.35.2.255 255.255.254.0 defines an address space of 26.35.2.0 - 26.35.3.255

The network ID would be: 26.35.2.0
The broadcast address would be 26.35.3.255
The host address range would be 26.35.2.1 - 26.35.3.254, which is inclusive of 26.35.2.255

Note: The address in question, is host assignable.
Note: The lesson is that a fourth octet of 255, does not always mean a broadcast address.


E.     17.35.36.0 255.255.254.0 defines an address space of 17.35.36.0 - 17.35.37.255

The network ID would be: 17.35.36.0
The broadcast address would be 17.35.37.255
The host address range would be 17.35.36.1 - 17.35.37.254, which is not inclusive of 17.35.36.0

Note: The address in question, happens to be a Network ID.

Best Regards,
Mike

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Nagaraja Thanthry
Cisco Employee
Cisco Employee

‎07-08-2010 08:10 PM

Hello,

The first one is invalid because all the host bits are zero's (last 9 bits are host bits for a 255.255.254.0 mask)

The second one is a valid address as the host bits will be 100000000 (This satisfies the condition that all host bits should not be zeros)

Third one is invalid because all host bits are one's

Fourth one is valid because the host bits are 011111111

Last one is invalid because again all the host bits are zero's.

Hope this helps.

Regards,

NT

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wang baojun
Level 1
Level 1
In response to Nagaraja Thanthry

‎07-08-2010 08:20 PM

hello ,

i have one question,, i donnot understand whether this ip add can assign to a host . please help

203.148.0.255 / 255.255.254.0

i know netmask 255.255.254.0   broadcast address is  203.148.1.255 . 

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Nagaraja Thanthry
Cisco Employee
Cisco Employee
In response to wang baojun

‎07-08-2010 08:26 PM

Hello,

If you have enabled "IP classless" feature, then you can assign that address. If not, that is an invalid address as 203.x.x.x is a Class C address and the default (classful) mask is 255.255.255.0

Hope this helps.

Regads,

NT

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wang baojun
Level 1
Level 1
In response to Nagaraja Thanthry

‎07-08-2010 08:39 PM

shf-gw1(config)#interface vlan1

shf-gw1(config-if)# ip address 203.148.0.255 255.255.255.0 secondary

Bad mask /24 for address 203.148.0.255

shf-gw1(config-if)# ip address 203.148.1.0 255.255.254.0 secondary 

shf-gw1(config-if)# ip address 203.148.0.255 255.255.254.0 secondary

shf-gw1(config-if)#     

shf-gw1(config-if)#

shf-gw1(config-if)#do ping 203.148.1.0 re 100

Type escape sequence to abort.

Sending 100, 100-byte ICMP Echos to 203.148.1.0, timeout is 2 seconds:

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Success rate is 100 percent (100/100), round-trip min/avg/max = 1/1/9 ms

shf-gw1(config-if)#do ping 203.148.0.255 re 100

Type escape sequence to abort.

Sending 100, 100-byte ICMP Echos to 203.148.0.255, timeout is 2 seconds:

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Success rate is 100 percent (100/100), round-trip min/avg/max = 1/1/9 ms

shf-gw1(config-if)#do show run interface vlan1

Building configuration...

Current configuration : 178 bytes

!

interface Vlan1

ip address 203.148.1.0 255.255.254.0 secondary

ip address 203.148.0.255 255.255.254.0 secondary

ip address 192.168.5.45 255.255.255.0

load-interval 30

end

shf-gw1(config-if)#

shf-gw1(config)# do ping 192.168.5.255

Type escape sequence to abort.

Sending 5, 100-byte ICMP Echos to 192.168.5.255, timeout is 2 seconds:

Reply to request 0 from 192.168.5.1, 16 ms

Reply to request 1 from 192.168.160.1, 1 ms

Reply to request 1 from 192.168.5.254, 1 ms

Reply to request 1 from 192.168.5.102, 1 ms

Reply to request 1 from 192.168.5.103, 1 ms

Reply to request 1 from 192.168.5.3, 1 ms

Reply to request 1 from 192.168.5.1, 1 ms

Reply to request 2 from 192.168.160.1, 1 ms

Reply to request 2 from 192.168.5.254, 1 ms

Reply to request 2 from 192.168.5.102, 1 ms

Reply to request 2 from 192.168.5.103, 1 ms

Reply to request 2 from 192.168.5.3, 1 ms

Reply to request 2 from 192.168.5.1, 1 ms

Reply to request 3 from 192.168.5.1, 1 ms

Reply to request 3 from 192.168.160.1, 1 ms

Reply to request 3 from 192.168.5.254, 1 ms

Reply to request 3 from 192.168.5.102, 1 ms

Reply to request 3 from 192.168.5.103, 1 ms

Reply to request 3 from 192.168.5.3, 1 ms

Reply to request 4 from 192.168.160.1, 1 ms

Reply to request 4 from 192.168.5.254, 1 ms

Reply to request 4 from 192.168.5.103, 1 ms

Reply to request 4 from 192.168.5.102, 1 ms

thanks NT , like this , i understood .. very clear

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yoda95823
Level 1
Level 1

‎07-11-2010 11:09 PM

Here are some sites to learn and practice subnetting:

http://easysubnet.com/
http://subnettingquestions.com

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michael.leblanc
Level 4
Level 4

‎07-13-2010 08:58 AM

rprudent:

The easiest way to get a handle on these, is with a thought process similar to the following:

  • With a mask of 255.255.254.0, it is the third octet that distinguishes adjacent Networks IDs.
  • If you subtract the value of the third octet (254) from 256 (0 - 255  represents 256 possible values), you will get 2.
  • This value (2) represents the increment between Network IDs.

E.g.: 113.10.4.0 255.255.254.0 would yield the following Network IDs:

113.10.0.0
113.10.2.0
113.10.4.0
113.10.6.0
.

. etc.
113.10.252.0
113.10.254.0


Note: that the third octet is incrementing by 2.

  • The address that follows a Network ID (e.g.: 113.10.4.0), is the first assignable host address (e.g.: 113.10.4.1).
  • The address that precedes the "next" Network ID (e.g.: 113.10. 6.0), would be the broadcast address (e.g.: 113.10.5.255).
  • The address that precedes the broadcast address (e.g.: 113.10.5.255), would be the last assignable host address (e.g.: 113.10.5.254).

Answer Assessment


A.     113.10.4.0 255.255.254.0 defines an address space of 113.10.4.0 - 113.10.5.255

The network ID would be: 113.10.4.0
The broadcast address would be 113.10.5.255
The host address range would be 113.10.4.1 - 113.10.5.254, which is not inclusive of 113.10.4.0

Note: The address in question, happens to be a Network ID.


B.     186.54.3.0 255.255.254.0 defines an address space of 186.54.2.0 - 186.54.3.255

The network ID would be: 186.54.2.0
The broadcast address would be 186.54.3.255
The host address range would be 186.54.2.1 - 186.54.3.254, which is inclusive of 186.54.3.0

Note: The address in question, is host assignable.


C.     175.33.3.255 255.255.254.0 defines an address space of 175.33.2.0 - 175.33.3.255

The network ID would be: 175.33.2.0
The broadcast address would be 175.33.3.255
The host address range would be 175.33.2.1 - 175.33.3.254, which is not inclusive of 175.33.3.255

Note: The address in question, happens to be the broadcast address.


D.     26.35.2.255 255.255.254.0 defines an address space of 26.35.2.0 - 26.35.3.255

The network ID would be: 26.35.2.0
The broadcast address would be 26.35.3.255
The host address range would be 26.35.2.1 - 26.35.3.254, which is inclusive of 26.35.2.255

Note: The address in question, is host assignable.
Note: The lesson is that a fourth octet of 255, does not always mean a broadcast address.


E.     17.35.36.0 255.255.254.0 defines an address space of 17.35.36.0 - 17.35.37.255

The network ID would be: 17.35.36.0
The broadcast address would be 17.35.37.255
The host address range would be 17.35.36.1 - 17.35.37.254, which is not inclusive of 17.35.36.0

Note: The address in question, happens to be a Network ID.

Best Regards,
Mike

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